73.CHEMICAL BONDING (20)- Covalent Bonding(19)- Bent’s rule(2).

After having learnt the statement of Bent’s rule in the earlier post , let us now study the rule with some examples.

We know that bond angle(θ) ∝ % s character ( from post no 71) .Then , according to Bent’s rule ,

737.jpg

Now let us apply this to various examples and study Bent’s rule . 

Example 1 – Methyl Fluoride (CH3F molecule) .

738.jpg

In a methyl fluoride molecule , fluorine is the most electronegative element. Just like methane (CH4) , this molecule is sp3 hybridised too.However, the distribution of s- character is not uniform in all hybrid orbitals and so bond angles change too.This change is due to the presence of F atom.  We can predict the change in bond angles using Bent’s rule.

According to Bent’s rule, the orbital with less %s character will be directed towards the more electronegative element. So, we can conclude that , the bond between C- F has less % s- character compared to C-H bonds. Also, the bond angle will decrease. Experimentally the H-C-F bond angle is found to be 108.2º , which is less than 109.5º .

When the %s character in C- F bond decreases, it is accounted for by increase in the  %s-character in the other three C-H bonds. So, C-H bonds have more %s character than expected ,resulting in increase in bond angle between them. Experimentally , the H-C-H bond is found to be 110.2º.

739.jpg

Example 2 –  Sulphur tetrafluoride (SF4) .

We were skeptical about the structure of this molecule in the earlier post.Let us study it’s structure using Bent’s rule. SF4  is sp3hybridized and has TBP geometry .There are four fluorine atoms and a lone pair in this molecule. As we know, according to Pauling scale of electronegativity , fluorine is the most electronegative element in the periodic table (refer post no 46) . In the TBP geometry , there are three equatorial positions and two axial positions.

7311.jpg

We can split sp3d as sp2 + pd. 

As seen above , the sp2 orbitals occupy the equatorial positions and the pd orbitals occupy  the axial position. How can we be sure of this ?  Let’s check whether this is true – 

Experimentally , we find out that the angle between equatorial substituents is 120º. We also know, 

i = – 1/ cosθ
∴i = – 1/ cos 120
i= –  1/ -(1/2) ..As cos 120 = -1/2
∴ i = 2
So, as the hybridization index is equal to 2 , the equatoriaorbitals are sp2 orbitals. 

Or we can use the formula from post 69 – fs=cosθ/(cosθ-1) 

∴fraction of s- character in equatorial positions with θ = 120º is ,

fs= cos(120) / (cos 120 -1)
= (-1/2) /(-1/2 -1)
=(-1/2)/ (-3/2)
= 1/3
=0.33.

 We know that it is sp2 orbitals which have 33% s- character , so the equatorial hybrid orbitals are sp2 hybridized.

lllly we know experimentally , that the angle with axial substituents is 90º. So, 

fs=cosθ/(cosθ-1)
       = cos 90 / cos90 – 1
= 0/(0-1)
=0.

So, the amount of s- character in axial orbitals is zero i.e the pd part(no s character)occupies the axial position. Applying Bent’s rule – 

7312.jpg

So the correct structure for this compound is  – 7313.jpg

We shall study some more examples of this rule in our next post. Till then , 

Be a perpetual student of life and keep learning …

Good day !

 

 

2 comments

  1. Thank you so much sir for explaining the Bent rule in such a simple way. I was having little difficulty in it, but after reading your post I developed interest in this topic and in inorganic chemistry because my interest was always been in organic chemistry.

    Like

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